Computer Science 9618/32 October/November 2021: question paper and mark scheme

Mark scheme

1(a)(i) One mark for each correct marking point (Max 2) 2 • 010111000110 (correct mantissa) • 0111 (correct exponent) 1(a)(ii) One mark for each correct consequence 2 One mark for each correct justification Consequence • The precision/accuracy of the number would be reduced Justification • … because the least significant bits of the original number have been truncated/lost // the original number had 13 bits / 14 bits with sign but the mantissa can only store 12 bits 1(b) One mark for each correct marking point (Max 3) 3 • To store the maximum range of numbers in the minimum number of bytes / bits • Normalisation minimises the number of leading zeros/ones represented • Maximising the number of significant bits // maximising the (potential) precision / accuracy of the number for the given number of bits • … enables very large / small numbers to be stored with accuracy. • Avoids the possibility of many numbers having multiple representations.

Mark scheme

2 One mark for each single correct line from Programming Paradigm to 4 Description Programming Paradigm Description Programs using the instruction set of a processor Declarative Programs based on events such as user actions or sensor outputs Imperative Programs using the concepts of class, inheritance, encapsulation and Low-level polymorphism Programs with an explicit sequence of commands that update the program Object oriented state, with or without procedure calls Programs that specify the desired result rather than how to get to it

Mark scheme

3(a) One mark for each marking point (Max 2) 2 • TYPE Parts = • (Monitor, CPU, SSD, HDD, LaserPrinter, Keyboard, Mouse) Complete answer TYPE Parts = (Monitor, CPU, SSD, HDD, LaserPrinter, Keyboard, Mouse) 3(b) One mark for each marking point (Max 2) 2 • TYPE SelectParts = ^ • correct data type chosen Parts Complete answer TYPE SelectParts = ^Parts

Mark scheme

4(a) One mark for each marking point (Max 2) 2 • <character>::= • $|%|&|*|# Complete answer <character>::= $|%|&|*|# 4(b)(i) For example: $A9E3 1 4(b)(ii) One mark for each marking point (Max 4) 4 • <password>::=<character> … • … <code> • <code>::= … • … <digit>|<capital_letter> • … |<digit><code>|<capital_letter><code> Complete answer <password>::=<character><code> <code>::=<digit>|<capital_letter>|<digit><code>|<capital_ letter><code>

Mark scheme

5(a) One mark for each correct marking point (Max 4) 4 • In both serial and sequential files records are stored one after the other … • … and need to be accessed one after the other • Serial files are stored in chronological order • Sequential files are stored with ordered records • … and stored in the order of the key field • In serial files, new records are added in the next available space / records are appended to the file • In sequential files, new records are inserted in the correct position. 5(b) Direct (access) 1 5(c) Sequential (access) 1

Mark scheme

6(a) One mark for each correct marking point (Max 5) 5 • A large message is divided up into a group of smaller chunks of the same size called packets • The packet has a header and a payload • The header contains a source IP address, destination IP address (and sequence number) • Each packet is dispatched independently • … and may travel along different routes / paths • The packets may arrive out of order • … and are reassembled into the original message at the destination • If packets are missing / corrupted a re-transmission request is sent. 6(b) One mark for each correct marking point (Max 3) 3 • The router examines the packet’s header • It reads the IP address of the destination (from the packet header) • A router has access to a routing table • …containing information about, e.g., available hops / netmask / gateway used • … and the status of the routes along the route • … the router decides on the next hop / best route • … and sends the packet on its next hop.

Mark scheme

7(a) One mark per two correct products (Max 3) 3 _ _ _ _ _ _ (Z =) A B C D + A B C D + A B C D + A B C D + _ ABCD+ABCD 7(b)(i) One mark for every two correct rows or columns (Max 2) 2 AB 00 01 11 10 00 0 0 1 0 01 0 0 1 1 CD 11 0 0 1 1 10 0 0 1 0 7(b)(ii) One mark for correct loop (Max 2) 2 AB 00 01 11 10 0 1 00 0 0 01 0 0 1 1 CD 11 0 0 1 1 01 0 0 1 0 7(b)(iii) One mark per correct marking point (Max 2) 2 • A B // A D • + A D // + A B (Z =) A B + A D // A D + A B 7(b)(iv) (Z =) A (B + D) // A (D + B) 1

Mark scheme

8(a) One mark for each correct marking point (Max 2) 2 • The SSL and TLS protocols provide communications security over the internet / network • … they provide encryption • They enable two parties to identify and authenticate each other • … and communicate with confidentiality and integrity. 8(b) One mark for each correct marking point (Max 4) 4 • An SSL/TLS connection is initiated by an application • … which becomes the client • The application which receives the connection becomes the server • Every new session begins with a handshake (as defined by the (SSL/TLS) protocols) • The client requests the digital certificate from the server // the server sends the digital certificate to the client • The client verifies the server’s digital certificate • …and obtains the server’s public key • The encryption algorithms are agreed • The symmetric • … session keys are generated / defined

Mark scheme

9(a)(i) One mark for correct statement (Max 1) 1 • Enables deep learning to take place • Where the problem you are trying to solve has a higher level of complexity it requires more layers to solve • To enable the neural network to learn and make decisions on its own • To improve the accuracy of the result. 9(a)(ii) One mark for each correct marking point (Max 4) 4 • Artificial neural networks are intended to replicate the way human brains work • Weights / values are assigned for each connection between nodes • The data are input at the input layer and are passed into the system • They are analysed at each subsequent (hidden) layer where characteristics are extracted / outputs are calculated • … this process of training / learning is repeated many times to achieve optimum outputs // reinforcement learning takes place • Decisions can be made without being specifically programmed • The deep learning net will have created complex feature detectors • The output layer provides the results • Back propagation (of errors) will be used to correct any errors that have been made. 9(b) One mark for each correct calculation as follows (Max 4) 5 • Node B (from Home) (Line 3 in table) • Node C (from Home) (Line 4 in table) • Node B and Node E (from A) (Lines 5 and 6 in table) • Node F and Node School (from E) (Lines 7 and 8 in table) • Node School (from F) (Line 9 in table) One mark for correct path (Max 1): • Home A E F School Node Cost from Home Heuristic Total Node (g) (h) (f = g + h) 1 Home 0 14 14 2 A 1 10 11 3 B 5 7 12 4 C 4 9 13 5 B 1+3=4 7 11 6 E 1+6=7 3 10 7 F 7+1=8 3 11 8 School 7 + 5 = 12 0 12 9 School 8 + 3 = 11 0 11 Final Path Home A E F School

Mark scheme

10(a) One mark for each correct marking point (Max 3) 3 • Must have a base case/stopping condition • Must have a general case • … which calls itself (recursively) // Defined in terms of itself • … which changes its state and moves towards the base case Unwinding can occur once the base case is reached. 10(b) One mark for each correct marking point (Max 3) 3 • A stack is a LIFO data structure • Each recursive call is pushed onto the stack • …. and is then popped as the function ends • Enables backtracking/unwinding … to maintain the required order. 10(c) One mark for each marking point (Max 2) 2 • Linked List • Queue Binary Tree 10(d) One mark for each marking point (Max 5) 5 • Checking if stack is full / empty using IF … THEN … (ELSE) … ENDIF • … correctly using StackFull() function • RETURN suitable message if stack is full • RETURN message if space available on stack • Incrementing TopOfStack pointer if space available • Assigning new data using correct NewInteger variable • … to correct the array element in ArrayStack[] array. Example algorithm FUNCTION AddInteger(NewInteger : INTEGER) RETURNS STRING IF StackFull() THEN RETURN "The stack is full" ELSE TopOfStack ← TopOfStack + 1 ArrayStack[TopOfStack] ← NewInteger RETURN "Item added" ENDIF ENDFUNCTION